Question

A column of mercury of length 10 cm is contained in the middle of a horizontal tube of length 1m which is closed at both the ends. The two equal lengths contain air at standard atmospheric pressure of 0.76 m of mercury. The tube is now turned to vertical position. By what approximate distance (in cm) will the column of mercury be displaced?

Answer 1

Let $Ac{m}^2$ be the area of cross-section of the tube. When the tube is in the horizontal position, as shown in the figure (a), the length of the air column on the left hand side region (I) and right hand side region (II) of the mercury column is 45 cm each, the total length of the tube being 100 cm with the mercury column length 10 cm in the middle. The pressure of air in each half is P = 76 cm of mercury and the volume of air in each half is $V=\left(45A\right)c{m}^3$.
When the tube is turned to vertical position, as shown in figure (b), the mercury column is displaced downwards. Let $xcm$ be the length of air column I, when the mercury column attains equillibrium. Volumn of column is $V_1=Axc{m}^3$ and let $P_1$ cm of Hg be the pressure of air in column I. The length of column II is $(90-x)cm$ and its volume is $V_2=(90-x)Ac{m}^3$. Let the pressure of air in column II be $P_{2}cm$ of Hg.

$\therefore P_1-P_2=10cm$ of Hg $\cdots \left(i\right)$
Initially (when the tube in the horizontal position) the pressure and volume of column I are P = 76 cm of Hg and V = 45A $c{m}^2$ respectively. Applying Boyle's law, we have, for column I.
$P_1{V}_1=PV$
or $P_1=\frac{PV}{V_1}=\frac{76cmHg\times 45Ac{m}^3}{Axc{m}^3}$
$P_1=\frac{76\times 45}{x}$ cm of Hg $\cdots \left(ii\right)$
Similarly, for column II, we have
$P_2{V}_2=PV$
or $P_2=\frac{PV}{V_2}=\frac{76cmofHg\times 45Ac{m}^3}{(90-x)Ac{m}^3}$
$=\frac{76\times 45}{(90-x)}$cm of Hg $\cdots \left(iii\right)$
Using equation (ii) and (iii) in (i), we get
$\frac{76\times 45}{x}-\frac{76\times 45}{(90-x)}=10$

which gives $x^2-774x+30780=0$
the two roots of this equation are x = 731.95 cm and 42.05 cm.

The first value 731.95 cm is obviously impossible.
Hence $x=42.05cm\approx 42cm$
$\therefore$ Displacement of mercury column
$=45cm-42cm=3cm$